3.357 \(\int \frac{1}{x^2 (a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{5 b}{a^3 \sqrt{a+b x}}-\frac{5 b}{3 a^2 (a+b x)^{3/2}}+\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{7/2}}-\frac{1}{a x (a+b x)^{3/2}} \]

[Out]

(-5*b)/(3*a^2*(a + b*x)^(3/2)) - 1/(a*x*(a + b*x)^(3/2)) - (5*b)/(a^3*Sqrt[a + b*x]) + (5*b*ArcTanh[Sqrt[a + b
*x]/Sqrt[a]])/a^(7/2)

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Rubi [A]  time = 0.0242279, antiderivative size = 80, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {51, 63, 208} \[ -\frac{5 \sqrt{a+b x}}{a^3 x}+\frac{10}{3 a^2 x \sqrt{a+b x}}+\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{7/2}}+\frac{2}{3 a x (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x)^(5/2)),x]

[Out]

2/(3*a*x*(a + b*x)^(3/2)) + 10/(3*a^2*x*Sqrt[a + b*x]) - (5*Sqrt[a + b*x])/(a^3*x) + (5*b*ArcTanh[Sqrt[a + b*x
]/Sqrt[a]])/a^(7/2)

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{x^2 (a+b x)^{5/2}} \, dx &=\frac{2}{3 a x (a+b x)^{3/2}}+\frac{5 \int \frac{1}{x^2 (a+b x)^{3/2}} \, dx}{3 a}\\ &=\frac{2}{3 a x (a+b x)^{3/2}}+\frac{10}{3 a^2 x \sqrt{a+b x}}+\frac{5 \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{a^2}\\ &=\frac{2}{3 a x (a+b x)^{3/2}}+\frac{10}{3 a^2 x \sqrt{a+b x}}-\frac{5 \sqrt{a+b x}}{a^3 x}-\frac{(5 b) \int \frac{1}{x \sqrt{a+b x}} \, dx}{2 a^3}\\ &=\frac{2}{3 a x (a+b x)^{3/2}}+\frac{10}{3 a^2 x \sqrt{a+b x}}-\frac{5 \sqrt{a+b x}}{a^3 x}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{a^3}\\ &=\frac{2}{3 a x (a+b x)^{3/2}}+\frac{10}{3 a^2 x \sqrt{a+b x}}-\frac{5 \sqrt{a+b x}}{a^3 x}+\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0065211, size = 33, normalized size = 0.45 \[ -\frac{2 b \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};\frac{b x}{a}+1\right )}{3 a^2 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x)^(5/2)),x]

[Out]

(-2*b*Hypergeometric2F1[-3/2, 2, -1/2, 1 + (b*x)/a])/(3*a^2*(a + b*x)^(3/2))

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Maple [A]  time = 0.013, size = 67, normalized size = 0.9 \begin{align*} 2\,b \left ( -{\frac{1}{{a}^{3}} \left ( 1/2\,{\frac{\sqrt{bx+a}}{bx}}-5/2\,{\frac{1}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-1/3\,{\frac{1}{{a}^{2} \left ( bx+a \right ) ^{3/2}}}-2\,{\frac{1}{\sqrt{bx+a}{a}^{3}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x+a)^(5/2),x)

[Out]

2*b*(-1/a^3*(1/2*(b*x+a)^(1/2)/b/x-5/2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/2))-1/3/a^2/(b*x+a)^(3/2)-2/a^3/(b*
x+a)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47718, size = 494, normalized size = 6.68 \begin{align*} \left [\frac{15 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt{a} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt{b x + a}}{6 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}, -\frac{15 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt{b x + a}}{3 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(15*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^2*
x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x), -1/3*(15*(b^3*x^3 + 2*a*b^2*x^2
+ a^2*b*x)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^2*x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x + a))/(a^4
*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)]

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Sympy [B]  time = 7.26392, size = 818, normalized size = 11.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x+a)**(5/2),x)

[Out]

-6*a**17*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4
) - 46*a**16*b*x*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b
**3*x**4) - 15*a**16*b*x*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2
)*b**3*x**4) + 30*a**16*b*x*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*
x**3 + 6*a**(33/2)*b**3*x**4) - 70*a**15*b**2*x**2*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a
**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 45*a**15*b**2*x**2*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x*
*2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**15*b**2*x**2*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2
)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 30*a**14*b**3*x**3*sqrt(1 + b*x/
a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 45*a**14*b**3*x**3
*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**14*
b**3*x**3*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)
*b**3*x**4) - 15*a**13*b**4*x**4*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*
a**(33/2)*b**3*x**4) + 30*a**13*b**4*x**4*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a
**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4)

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Giac [A]  time = 1.19638, size = 88, normalized size = 1.19 \begin{align*} -\frac{5 \, b \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} - \frac{2 \,{\left (6 \,{\left (b x + a\right )} b + a b\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}} - \frac{\sqrt{b x + a}}{a^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

-5*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) - 2/3*(6*(b*x + a)*b + a*b)/((b*x + a)^(3/2)*a^3) - sqrt(b*
x + a)/(a^3*x)